Direction of the Electric Field

AI Thread Summary
The direction of the electric field (E) is determined by the nature of the charge involved; positive charges create an outward field, while negative charges create an inward field. In this case, the electric field is assumed to point upward based on the force experienced by a negative charge, which is attracted to a positive charge. The relationship between force (F), electric field (E), and charge (q) is described by the equation F = E x q, where the direction of F indicates the direction of E when considering the sign of the charge. The electric field can vary depending on external influences from other charges, but for this scenario, it is safe to conclude that the electric field points upward. Understanding these principles clarifies the behavior of electric fields in relation to charged particles.
Raybulous
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Homework Statement


a.jpg


Homework Equations


F = E x q

The Attempt at a Solution


I was told the direction of the E field should be pointing upwards but I have no clue as to why. I found the Force = 0.0012 N but not the direction.
 
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Hi, I am not sure if anybody had ever told you this, but if you take law of elementary charges e.g. like charges repel and opposite attracts; which is basically indicates the direction of the attraction or repulsion force is felt, And electric field is just the opposite of it which is again said as, q+ has E outwards and q- has inward electric field. Therefore, you can see that the negative q or electron is pulled or attracted to another charge by 0.0012N in the direction of F but as it's Electric field is drawn to it because it is a negative charge, so Electric field is actually pointing upward which is where you electron is.
 
jackMybrain@ru said:
Hi, I am not sure if anybody had ever told you this, but if you take law of elementary charges e.g. like charges repel and opposite attracts; which is basically indicates the direction of the attraction or repulsion force is felt, And electric field is just the opposite of it which is again said as, q+ has E outwards and q- has inward electric field. Therefore, you can see that the negative q or electron is pulled or attracted to another charge by 0.0012N in the direction of F but as it's Electric field is drawn to it because it is a negative charge, so Electric field is actually pointing upward which is where you electron is.
So to find the electric field direction we only have to look at the charge of the particle?
Hence the electric field is pointing towards the particle and not flowing in a constant direction?
 
It's flowing in a constant direction but it also changes depending on if there are any other sources e.g. outer electric field line exerted from a differently charged particle or if your charge particle is placed inside another electric field, which will change it's electric field direction. So, can't say directly unless you have made the question a bit more specific with condition. But for this case, it's safe to assume that Electric field is going upward by looking at the direction of the force and assuming that there is no repelling force by an unknown charge on the top y-axis.
 
Raybulous said:

Homework Statement


View attachment 87949

Homework Equations


F = E x q

The Attempt at a Solution


I was told the direction of the E field should be pointing upwards but I have no clue as to why. I found the Force = 0.0012 N but not the direction.
Look at the equation F=E*q. F and E are vectors, e is a scalar, it is negative now. What happens to the direction of a vector if you multiply it with a negative scalar? If the result, F, points downward, what is the direction of E?
 
Raybulous said:
So to find the electric field direction we only have to look at the charge of the particle?
Hence the electric field is pointing towards the particle and not flowing in a constant direction?
A charge does not feel its own electric field. When we say that a charged particle is placed in an electric field E, and a force acts on it, the force felt by the charge is F=qE, parallel or antiparallel with E, depending on the sign of the charge.
The net field will be modified by the added charge, but usually it is assumed that the charge is small and it disturbs the original field only very near to itself.
 
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